What is the extraneous solution to these equations? $\dfrac{x^2 - 55}{x + 3} = \dfrac{26}{x + 3}$
Multiply both sides by $x + 3$ $ \dfrac{x^2 - 55}{x + 3} (x + 3) = \dfrac{26}{x + 3} (x + 3)$ $ x^2 - 55 = 26$ Subtract $26$ from both sides: $ x^2 - 55 - (26) = 26 - (26)$ $ x^2 - 55 - 26 = 0$ $ x^2 - 81 = 0$ Factor the expression: $ (x - 9)(x + 9) = 0$ Therefore $x = 9$ or $x = -9$ The original expression is defined at $x = 9$ and $x = -9$, so there are no extraneous solutions.